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complyorexplain
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Trading puzzle

July 16th, 2020, 1:17 pm

Image
See the chart above, and start with the red line. We buy a quantity X at a price of 12, then sell when the price returns to 10, and so on. Each time we lose the change in price times the amount held = X(12-10) = 2X, which happens 4 times, so the total loss is 4 x 2X = 8X.
 
Then follow the blue line, where the price change is twice as frequent. This time there are 8 changes, the amount lost each time is X(11-10) = X, so the total loss is 8X, the same as before. Then suppose we halve the time, and the change in price again (not shown on the chart). It is easy to show that the loss is always 8X or (which is the same thing) the loss per unit time the same.
 
Then suppose we make infinitely many trades in the same time. Is the loss still 8X? For in the infinite case, there is no change in price, so it is impossible that there can be any loss per unit time.
 
Or can we say that there are infinitely many trades, each with an infinitesimally small loss?
 
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katastrofa
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Re: Trading puzzle

July 16th, 2020, 5:17 pm

So you're saying that [$]2/2^\infty \cdot 2^\infty \cdot 4 \neq 8[$]?
OK.

Seriously, sounds like Kolmogorov vs de Finetti again (sigma additivity) :-) I think quartz has better credentials to answer.
 
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complyorexplain
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Re: Trading puzzle

July 17th, 2020, 8:15 am

I am posting this because someone else expressed a doubt. 
 
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Alan
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Re: Trading puzzle

July 17th, 2020, 4:09 pm

[$]\int_0^1 dx = 1[$]
 
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katastrofa
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Re: Trading puzzle

July 17th, 2020, 5:06 pm

Isn't the problem analogous to de Finetti Lottery?
 
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Alan
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Re: Trading puzzle

July 17th, 2020, 5:29 pm

Curiously, I just started reading de Finetti's book, but am still at the introduction.  :D

But I googled the concept. I don't see any probability in the OP's setup. 

He or she seems to be simply partitioning a fixed interval (the total loss) into N segments (the loss per trade), with each individual loss [$]\Delta x[$] proportional to 1/N. Taking the simplest example of a fixed interval = total loss = 1, and so [$]\Delta x = 1/N[$], we have

[$] \lim_{N \rightarrow \infty} \sum_{i=1}^N \Delta x = \lim_{N \rightarrow \infty} \sum_{i=1}^N \frac{1}{N} = 1[$]

or, in modern notation:

[$] \int_0^1 dx = 1[$]. 
 
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katastrofa
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Re: Trading puzzle

July 17th, 2020, 6:51 pm

OK, thanks. It just proves how I don't get HFT :-)
 
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complyorexplain
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Re: Trading puzzle

July 17th, 2020, 10:26 pm

But I googled the concept. I don't see any probability in the OP's setup. 
Yes correct. Nothing to do with probability. The puzzle is that at the limit the price never changes. So how can any profit or loss be made.
 
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complyorexplain
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Re: Trading puzzle

July 17th, 2020, 10:49 pm

Henri Léon Lebesgue (1875-1941), Measure and the Integral, edited with a biographical essay by Kenneth Ownsworth May (1915-1977), The Mathesis Series, Holden-Day, 1966, xii + 194 pages.

Formerly, when I was a schoolboy, the teachers and pupils had been satisfied with this reasoning by passage to the limit. However, it ceased to satisfy me when some of my schoolmates showed me, along about my fifteenth year, that one side of a triangle is equal to the sum of the other two and that π=2. Suppose that ABC is an equilateral triangle and that D, E, and F are the midpoints of BA, BC, and CA. The length of the broken line [= polygonal path] BDEFC is AB+AC. If we repeat this procedure with the triangles DBE and FEC, we get a broken line of the same length made up of eight segments, etc. Now these broken lines have BC as their limit, and hence the limit of their lengths, that is, their common length AB+AC, is equal to BC. The reasoning with regard to π is analogous.

Nothing, absolutely nothing, distinguishes this reasoning from what we used to evaluate the circumference and area of a circle, the surface and volume of a cylinder, a cone, and a sphere. This result has been extremely instructive to me.

Besides, every paradox is highly instructive. In my opinion, the critical examination of paradoxes and the correction of erroneous reasoning should be standard exercises, frequently repeated at the secondary level.
 
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katastrofa
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Re: Trading puzzle

July 18th, 2020, 9:41 am

De Finetti is also about the infinite countability which he argued was impractical (he's been mentioned in the neighbouring thread, hence my thoughts went there).
Here the "paradox" is about interchanging the limit and the sum, I think.
 
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Alan
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Re: Trading puzzle

July 18th, 2020, 2:31 pm

But I googled the concept. I don't see any probability in the OP's setup. 
Yes correct. Nothing to do with probability. The puzzle is that at the limit the price never changes. So how can any profit or loss be made.

It's too trivial to be called a paradox. 
While it's true that the "price change" [$]\frac{1}{N} \rightarrow 0[$], 
it's also true that "number of trades" x "price change" = [$]N \times \frac{1}{N} = 1[$]. 
End of story.
Why is that so difficult!? 
This whole thread is trollish.  :(
 
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katastrofa
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Re: Trading puzzle

July 18th, 2020, 3:05 pm

Yeah, it's not a paradox (but not a good troll either - I'd give it 1/10).
It's an error - an invalid interchanging the limit and summation: [$] \lim_{N \rightarrow \infty} \sum_{i=1}^N \frac{1}{N} \neq \sum_{i=1}^\infty \lim_{N \rightarrow \infty} \frac{1}{N}[$]
 
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complyorexplain
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Re: Trading puzzle

July 19th, 2020, 10:30 am

It was Henri Lebesgue who called it a paradox, not I!

The reason I asked here was that we have taken this 'puzzle' to two separate mathematicians, one of them eminent by any standard, but they disagree on the result. No intention to troll, apologies.
 
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Collector
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Re: Trading puzzle

July 19th, 2020, 2:12 pm

it would take a lot of time for infinite number of trades even at NASDAQ's nano-second time stamp, prepare for gray hair! (ohh yes now a troll on the thread, sorry), but yes what we can do with infinite and zero always fascinating...
 
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katastrofa
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Re: Trading puzzle

July 19th, 2020, 9:54 pm

Hey, Alan's remark about trolling referred to my posts - I hope so! :-)

This is an interesting problem. Now that you rang a bell (Lebesgue), what if I tell Alan that the function I obtain by constantly dividing the trade frequency has the feature of the Dirichlet function, namely after infinitely many divisions it has a non-zero value on a rational part of its domain, and is 0 on the non-rational part. It's nowhere continuous and hence non-integrable in the Riemann sense? :-D </end of troll>

Another thing is the difference between the generating process and the observation, as noticed above. One can define a process, which is an infinite series of events, but in practice you can perform the measurement only on a subset of them. If you have zero measurements - you know nothing, if you know the whole process - you don't need to estimate, if you're inbetween - you have the Bayesian inference.
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